Let $f(x, y) = \sin(xy)$. Suppose $\vec{a} = (4, 0)$ and $\vec{v} = \left( 0, 1 \right)$. Find the directional derivative of $f(x, y)$ at $\vec{a}$ in the direction of $\vec{v}$. $\dfrac{\partial f}{\partial v} = $
Answer: When a directional derivative is in the direction $(1, 0)$, $(0, 1)$, $(-1, 0)$, or $(0, -1)$, it becomes a regular partial derivative. Because $v = (0, 1)$, the directional derivative we want to find is also $\dfrac{\partial f}{\partial y}$ evaluated at $(4, 0)$. $\begin{aligned} &\dfrac{\partial f}{\partial y} = x\cos(xy) \\ \\ &\dfrac{\partial f}{\partial y}(4, 0) = 4 \end{aligned}$ In conclusion, the directional derivative of $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $4$.